Video solution:

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Written solution:

Let $m=2^{101}+2^{51}+1$. We factor the numerator around this divisor: $$2^{202}+202=(2^{101}+1)^2-(2^{51})^2+201.$$ By the difference of squares, $$(2^{101}+1)^2-(2^{51})^2=(2^{101}+2^{51}+1)(2^{101}-2^{51}+1),$$ which is a multiple of $m$. Therefore $$2^{202}+202\equiv 201\pmod m.$$

Thus, the correct answer is D .